[proofplan]
We prove uniqueness by contradiction through the midpoint argument. If two distinct elements of $V$ attain the same minimal error, then their midpoint also lies in $V$ because $V$ is a linear subspace. After normalising the two error vectors, strict convexity forces the midpoint error to have strictly smaller norm, contradicting minimality. The zero-error case is separated first because normalisation by the error is then not available.
[/proofplan]
[step:Compare an arbitrary second minimizer with the given minimizer]
Let $w \in V$ be any element satisfying
\begin{align*}
\|x-w\|_X = \inf_{v \in V}\|x-v\|_X.
\end{align*}
Define the minimal error $E \in [0,\infty)$ by
\begin{align*}
E := \inf_{v \in V}\|x-v\|_X.
\end{align*}
Then
\begin{align*}
\|x-v^*\|_X = E
\end{align*}
and
\begin{align*}
\|x-w\|_X = E.
\end{align*}
It is enough to prove $w=v^*$, since $w$ was an arbitrary minimizer.
[/step]
[step:Handle the zero-error case before normalising]
Assume first that $E=0$. Since $\|x-v^*\|_X=0$, the definiteness of the norm gives $x-v^*=0$, so $x=v^*$. Likewise $\|x-w\|_X=0$ gives $x=w$. Therefore $w=v^*$.
[/step]
[step:Use strict convexity on the normalized error vectors]
Assume now that $E>0$. Define the normalized error vectors $a,b \in X$ by
\begin{align*}
a := \frac{x-v^*}{E}
\end{align*}
and
\begin{align*}
b := \frac{x-w}{E}.
\end{align*}
By homogeneity of the norm,
\begin{align*}
\|a\|_X = \frac{\|x-v^*\|_X}{E} = 1
\end{align*}
and
\begin{align*}
\|b\|_X = \frac{\|x-w\|_X}{E} = 1.
\end{align*}
Suppose, for contradiction, that $w \ne v^*$. Then $a \ne b$, because $a=b$ would imply $x-v^*=x-w$ and hence $v^*=w$. Since $X$ is strictly convex and $a,b$ are distinct unit vectors, we have
\begin{align*}
\left\|\frac{a+b}{2}\right\|_X < 1.
\end{align*}
[guided]
We are in the case $E>0$, so division by $E$ is legitimate. The purpose of normalising is to place the two error vectors on the unit sphere, because strict convexity is a statement about distinct unit vectors.
Define $a,b \in X$ by
\begin{align*}
a := \frac{x-v^*}{E}
\end{align*}
and
\begin{align*}
b := \frac{x-w}{E}.
\end{align*}
Since $v^*$ and $w$ both attain the same minimal error $E$, homogeneity of the norm gives
\begin{align*}
\|a\|_X = \left\|\frac{x-v^*}{E}\right\|_X = \frac{1}{E}\|x-v^*\|_X = 1
\end{align*}
and
\begin{align*}
\|b\|_X = \left\|\frac{x-w}{E}\right\|_X = \frac{1}{E}\|x-w\|_X = 1.
\end{align*}
Now assume, toward a contradiction, that $w \ne v^*$. Then the normalized error vectors are distinct. Indeed, if $a=b$, then multiplying by $E$ gives $x-v^*=x-w$, and subtracting $x$ from both sides gives $v^*=w$, contrary to the assumption.
Thus $a$ and $b$ are distinct unit vectors. By strict convexity of $(X,\|\cdot\|_X)$, the midpoint of any two distinct unit vectors has norm strictly less than $1$. Applying this definition to $a$ and $b$ yields
\begin{align*}
\left\|\frac{a+b}{2}\right\|_X < 1.
\end{align*}
This is the only point where strict convexity is used: ordinary convexity would only give a non-strict inequality, which would not contradict minimality.
[/guided]
[/step]
[step:Convert the strict midpoint inequality into a better approximation]
Define the midpoint approximant $m \in X$ by
\begin{align*}
m := \frac{v^*+w}{2}.
\end{align*}
Since $V$ is a linear subspace and $v^*,w \in V$, we have $m \in V$. Also,
\begin{align*}
x-m = x-\frac{v^*+w}{2} = \frac{x-v^*}{2}+\frac{x-w}{2} = E\frac{a+b}{2}.
\end{align*}
Therefore, by homogeneity of the norm and the strict midpoint inequality,
\begin{align*}
\|x-m\|_X = E\left\|\frac{a+b}{2}\right\|_X < E.
\end{align*}
This contradicts the definition of $E$ as the infimum of $\|x-v\|_X$ over all $v \in V$, because $m \in V$.
[/step]
[step:Conclude that every minimizer equals the given minimizer]
The contradiction shows that the assumption $w \ne v^*$ is impossible when $E>0$. Together with the zero-error case, every $w \in V$ satisfying
\begin{align*}
\|x-w\|_X = \inf_{v \in V}\|x-v\|_X
\end{align*}
must equal $v^*$. Hence $v^*$ is the unique best approximation to $x$ from $V$.
[/step]
